Let the tangents drawn from the origin to the circle, $x^{2}+y^{2}-8 x-4 y+16=0$ touch it at the points $A$ and $B .$ The $(A B)^{2}$ is equal to

If the tangent at the point $P$ on the circle ${x^2} + {y^2} + 6x + 6y = 2$ meets the straight line $5x - 2y + 6 = 0$ at a point $Q$ on the $y$- axis, then the length of $PQ$ is

The equation of the normal to the circle ${x^2} + {y^2} = 9$ at the point $\left( {\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right)$ is

Tangents are drawn from any point on the circle $x^2 + y^2 = R^2$ to the circle $x^2 + y^2 = r^2$. If the line joining the points of intersection of these tangents with the first circle also touch the second, then $R$ equals

If a circle, whose centre is $(-1, 1)$ touches the straight line $x + 2y + 12 = 0$, then the coordinates of the point of contact are

A circle with centre $(a, b)$ passes through the origin. The equation of the tangent to the circle at the origin is